Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Apr 17th 2013, 05:44 PM
Fratricide
1) i: Find the value of c such that y = x + c is a tangent to the parabola y = x2 - x - 12.
ii: Find the values of m for which the straight line y = mx + 6 is tangent to the parabola y = -2x2 - 6x + 2.

2) Determine the equation of the following parabola:

Attachment 27999

• Apr 17th 2013, 05:47 PM
Prove It
What have you tried so far?
• Apr 17th 2013, 08:56 PM
Fratricide
1. x2 - x - 12 = x +c
x2 - 2x - 12 = c

What value would I substitute for x? I think it has something to do with the equations being tangental, but I can't remember exactly.

2. I know that I have to substitute coordinates into the equation ax2 + bx + c and solve simultaneously, but I don't know how to solve simultaneous equations with three variables.
• Apr 17th 2013, 10:22 PM
ibdutt
Now we have the curve y = x^2 - x -12. Slope of tangent to the curve is given by dy/dx. for this curve dy/dx = 2x -1 Let the point of contact be (h,k)
Thus slope of tangent = 2h-1. The slope of the given line y = x + c is 1. Hence 2h - 1 = 1 that gives h = 1. Form the equation of the curve , by pugging in value of h=1 for x we get ythe value of k since (h,k) lies on the curve.
That gives k = 12. Hence the point of contact is ( 1, - 12 ) and slope of tangent is = 1. Thus its equation is y + 12 = 1 ( x - 1 ) that is y = x - 13. Comparing it with the given equation we get c = -13.
• Apr 18th 2013, 12:52 AM
Fratricide
Quote:

Originally Posted by ibdutt
Slope of tangent to the curve is given by dy/dx. for this curve dy/dx = 2x -1

I don't think I quite understand? I got the rest of your post, but this doesn't fully make sense.
• Apr 18th 2013, 01:06 AM
MarkFL
Quote:

Originally Posted by Fratricide
I don't think I quite understand? I got the rest of your post, but this doesn't fully make sense.

He's using calculus, but we may use a pre-calculus technique instead, since you posted in the Algebra forum. You correctly found:

$\displaystyle x^2-2x-12=c$

If we arrange this in standard quadratic form, we get:

$\displaystyle x^2-2x-(12+c)=0$

Now, if the line and parabola are tangent, this means they share only one point in common, i.e., the above quadratic has only one root. What condition on a quadratic must hold in order for there to be only one root?
• Apr 18th 2013, 02:06 AM
Fratricide
The discriminant must be equal to zero?
• Apr 18th 2013, 02:08 AM
MarkFL
Yes, good! Now what is the discriminant, and what do you find when you equate it to zero?
• Apr 18th 2013, 02:13 AM
Fratricide
Is - (12 + c) the c value for the discriminant?
• Apr 18th 2013, 02:20 AM
MarkFL
Yes, it is the constant term. :D
• Apr 18th 2013, 02:45 AM
Fratricide
Alright, thanks. I'll have a go solving it tomorrow -- it's getting late here.

This may sound ridiculously stupid, but why does the equation become x^2 - 2x - (12 + c) = 0 as opposed to something like, say, x^2 - 2x + c -12 = 0? Specifically, why does the c get tied to the 12 when it's moved?
• Apr 18th 2013, 02:51 AM
MarkFL
Since c is a constant, it should be grouped with the constant 12, as the other two terms have a power of x as factors. Since it began with a positive sign on the right, when we move it to the left, we are essentially subtracting c from both sides.
• Apr 18th 2013, 07:27 PM
jpritch422
This is the solution to part ii of problem 1:

Since the tangent line and the given curve have a point in common we have:

$\displaystyle mx + 6 = -2x^2 - 6x + 2$

$\displaystyle 2x^2 + (6 + m)x + 4 = 0$

slope of $\displaystyle m = \displaystyle\frac{dy}{dx}$ where $\displaystyle \displaystyle\frac{dy}{dx} = \displaystyle\frac{d}{dx} (-2x^2 - 6x + 2) = -4x - 6 = 0$

so $\displaystyle 6 + m = 6 - 4x - 6 = -4x$

$\displaystyle 2x^2 -4x^2 + 4 = 0$

$\displaystyle -2x^2 + 4 = 0$; $\displaystyle x^2 = 2$; $\displaystyle x = \pm \sqrt{2}$

if $\displaystyle x = \sqrt{2}$; $\displaystyle m = -4\sqrt{2} - 6 \approx -11.657$

if $\displaystyle x = -\sqrt{2}$; $\displaystyle m = 4\sqrt{2} - 6 \approx -.343$
• Apr 18th 2013, 07:46 PM
MarkFL